Question Number 194304 by 281981 last updated on 02/Jul/23 Answered by cortano12 last updated on 03/Jul/23 $$\:\:\:\:\mathrm{sin}\:\left(\mathrm{A}+\mathrm{B}\right)\:\mathrm{cos}\:\left(\mathrm{A}+\mathrm{B}\right)\:\mathrm{sin}\:\left(\mathrm{A}−\mathrm{B}\right)\:\mathrm{cos}\:\left(\mathrm{A}−\mathrm{B}\right)=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\:\:\:\mathrm{sin}\:\left(\mathrm{2A}+\mathrm{2B}\right)\:\mathrm{sin}\:\left(\mathrm{2A}−\mathrm{2B}\right)=\:\mathrm{2} \\ $$$$\:\:\:\underbrace{\boldsymbol{{A}}} \\ $$ Commented by…
Question Number 194143 by mathbd last updated on 28/Jun/23 Commented by a.lgnaoui last updated on 29/Jun/23 $$\mathrm{H}? \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 194252 by horsebrand11 last updated on 01/Jul/23 $$\:\:\mathrm{Find}\:\mathrm{V}\:=\:\mathrm{tan}\:\mathrm{9}°−\mathrm{tan}\:\mathrm{27}°−\mathrm{tan}\:\mathrm{63}°+\mathrm{tan}\:\mathrm{81}° \\ $$ Answered by peter frank last updated on 01/Jul/23 $$\mathrm{Qn}\:\mathrm{177542} \\ $$ Answered by…
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Question Number 193749 by Mingma last updated on 19/Jun/23 Answered by som(math1967) last updated on 19/Jun/23 Commented by som(math1967) last updated on 19/Jun/23 $${AD}={DC}={AC}=\mathrm{12}{cm} \\…
Question Number 193691 by cortano12 last updated on 18/Jun/23 $$ \\ $$ Commented by cortano12 last updated on 18/Jun/23 Answered by MM42 last updated on…
Question Number 193595 by Rupesh123 last updated on 17/Jun/23 Answered by MM42 last updated on 17/Jun/23 $${sin}^{\mathrm{2}} \alpha+{sin}^{\mathrm{2}} \beta−{sin}\alpha{cos}\beta−{sim}\beta{cos}\alpha=\mathrm{0} \\ $$$$\Rightarrow{sin}\alpha\left({sin}\alpha−{cos}\beta\right)+{sin}\beta\left({sin}\beta−{cos}\alpha\right)=\mathrm{0} \\ $$$${sin}\alpha\left(\mathrm{2}{sin}\left(\frac{\alpha+\beta}{\mathrm{2}}−\frac{\pi}{\mathrm{4}}\right){cos}\left(\frac{\alpha−\beta}{\mathrm{2}}+\frac{\pi}{\mathrm{4}}\right)\right)+{sin}\beta\left(\mathrm{2}{sin}\left(\frac{\alpha+\beta}{\mathrm{2}}−\frac{\pi}{\mathrm{4}}\right){cos}\left(\frac{\beta−\alpha}{\mathrm{2}}+\frac{\pi}{\mathrm{4}}\right)\right)=\mathrm{0} \\ $$$$\Rightarrow{sin}\left(\frac{\alpha+\beta}{\mathrm{2}}−\frac{\pi}{\mathrm{4}}\right)\left({sin}\alpha{cos}\left(\frac{\alpha−\beta}{\mathrm{2}}+\frac{\pi}{\mathrm{4}}\right)+{sin}\beta{cos}\left(\frac{\beta−\alpha}{\mathrm{2}}+\frac{\pi}{\mathrm{4}}\right)\right)=\mathrm{0}…
Question Number 193646 by Shlock last updated on 17/Jun/23 Answered by som(math1967) last updated on 17/Jun/23 $$\:\left({cos}^{\mathrm{2}} {x}+{sin}^{\mathrm{2}} {x}\right)^{\mathrm{2}} −\mathrm{2}{sin}^{\mathrm{2}} {xcos}^{\mathrm{2}} {x} \\ $$$$=\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{2}{sinxcosx}\right)^{\mathrm{2}} \\…
Question Number 193622 by a.lgnaoui last updated on 17/Jun/23 $$\mathrm{determiner}\:\mathrm{l}\:\mathrm{angle}\:\boldsymbol{\mathrm{x}} \\ $$ Commented by a.lgnaoui last updated on 17/Jun/23 Answered by cherokeesay last updated on…
Question Number 193609 by Mingma last updated on 17/Jun/23 Answered by Subhi last updated on 17/Jun/23 $$\frac{{x}}{{sin}\left(\mathrm{60}\right)}=\frac{{z}}{{sin}\left(\mathrm{120}−{y}\right)}\: \\ $$$$\frac{{z}}{{sin}\left(\mathrm{60}\right)}=\frac{\mathrm{1}}{{sin}\left(\mathrm{60}−{y}\right)}\:\Rrightarrow\:{z}=\frac{{sin}\left(\mathrm{60}\right)}{{sin}\left(\mathrm{60}−{y}\right)} \\ $$$${x}=\frac{{sin}^{\mathrm{2}} \left(\mathrm{60}\right)}{{sin}\left(\mathrm{60}−{y}\right){sin}\left(\mathrm{120}−{y}\right)} \\ $$$$\frac{{x}+{z}}{{sin}\left(\mathrm{120}\right)}=\frac{{z}}{{sin}\left(\mathrm{60}−{y}\right)} \\…